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Conversion of Binary, Octal and Hexadecimal Numbers From Binary to Octal Starting at the binary point and working left, separate the bits into groups of three and replace each group with the corresponding octal digit. 100010112 = 010 001 011 = 2138 From Binary to Hexadecimal Starting at the binary point and working left, separate the bits into groups of four and replace each group with the corresponding hexadecimal digit. 100010112 = 1000 1011 = 8B16 From Octal to Binary Replace each octal digit with the corresponding 3-bit binary string. 2138 = 010 001 011 = 100010112 From Hexadecimal to Binary Replace each hexadecimal digit with the corresponding 4-bit binary string. 8B16 = 1000 1011 = 100010112 Conversion of Decimal Numbers From Decimal to Binary 2 139 2 69 1 1 2 34 0 2 17 1 2 8 0 2 4 0 2 2 0 1 LSD 13910 = 100010112 MSD From Binary to Decimal 100010112 = 1×27 + 0×26 + 0×25 + 0×24 + 1×23 + 0×22 + 1×21 + 1×20 = 128 + 8 + 2 + 1 Conversion of Fractions Starting at the binary point, group the binary digits that lie to the right into groups of three or four. 0.101112 = 0.101 110 = 0.568 0.101112 = 0.1011 1000 = 0.B816 Problems Convert the following Binary 10011010 Octal Decimal Hex 2705 2705 3BC Binary 10011010 10111000101 101010010001 1110111100 Octal 232 2705 5221 1674 8 2705 1 8 338 2 8 42 2 5 Decimal 154 1477 2705 956 16 2705 1 16 169 9 10=A Hex 9A 5C5 A91 3BC Add 1 1 1 1 + 1 0 0 1 + 1 0 0 0 1 1 0 0 0 1 1 1 1 1 0 0 1 1 1 0 0 1 1 1 1 1 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 0 0 Subtract 1 - - Multiply normally for implementation - add the shifted multiplicands one at a time. 1 1 1 0 = 14 1 1 0 1 = 13 1 1 1 0 0 0 0 0 1 1 1 0 + 1 1 1 0 1 1 1 0 * 0 1 1 1 1 1 0 1 1 0 1 1 1 1 0 + 0 0 0 0 0 1 1 1 0 + 1 1 1 0 1 0 0 0 1 1 0 + 1 1 1 0 1 1 1 0 1 1 0 * 0 0 (8 bits) Divide 1101 110 1111) 11000101| 1111 | 1001101| 1111 | 1101) 1011001| 1101 | 100101| 1101 | 10001| 1011| 0000 0000| | 10001| 1111| 10 1001 1101) 1111001| 1101 | 10001| 0000 | 10001| 0000 | 10001| 1101| 100 1011 Sign-Magnitude 0 = positive 1 = negative n bit range = -(2n-1-1) to +(2n-1-1) 4 bits range = -7 to +7 2 possible representation of zero. 2's Complement flip bits and add one. n bit range = -(2n-1) to +(2n-1-1) 4 bits range = -8 to +7 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 =0 =1 =2 =7 = -8 = -7 = -2 = -1 Example 1110 0001 0010 = 14 flip bits add one WRONG this is not -14. Out of range. Need 5 bits 0 1 1 1 0 = 14 1 0 0 0 1 flip bits 1 0 0 1 0 add one. This is -14. Sign Extend add 0 for positive numbers add 1 for negative numbers Add 2's Complement 1110 = -2 + 1101 11011 = -3 1110 = -2 + 0011 =3 ignore carry = -5 10001 ignore carry = 1 Be careful of overflow errors. An addition overflow occurs whenever the sign of the sum if different from the signs of both operands. Ex. 0100 =4 1100 = -4 + 0101 =5 + 1011 = -5 1001 = -7 WRONG 10111 ignore carry = 7 WRONG Multiply 2's Complement * 1110 = -2 1101 = -3 11111110 + + = -2 0011 =3 11111110 0000000 +1 1 1 1 1 1 0 11111110 111111010 111110 111110110 + sign extend to 8 bits * 1110 00010 100000110 * ignore carry negate -2 for sign bit ignore carry = 6 10010 = -14 10011 = -13 1111110010 sign extend to 10 bits + 111110010 11111010110 ignore carry + 00000000 1111010110 + 0000000 1111010110 + 001110 10010110110 negate -14 for sign bit ignore carry = 182 sign extend to 8 bits ignore carry = -6 Floating-Point Numbers mantissa x (radix)exponent The floating-point representation always gives us more range and less precision than the fixed-point representation when using the SAME number of digits. Mantissa Sign sign exponent General format 0 1 9 Mantissa 8-bit excess-127 sign characteristic 32-bit standard Mantissa magnitude 31 23-bit normalized fraction Implied binary point 0 1 12 63 Mantissa 11-bit excess 52-bit normalized fraction sign 1023 charactstic 64-bit standard Normalized fraction - the fraction always starts with a nonzero bit. e.g. 0.01 … x 2e would be normalized to 0.1 … x 2e-1 1.01 … x 2e would be normalized to 0.101 … x 2e+1 Since the only nonzero bit is 1, it is usually omitted in all computers today. Thus, the 23-bit normalized fraction in reality has 24 bits. The exponent is represented in a biased form. • • • • If we take an m-bit exponent, there are 2m possible unsigned integer values. Re-label these numbers: 0 to 2m-1 → -2m-1 to 2m-1-1 by subtracting a constant value (or bias) of 2m-1 (or sometimes 2m-1-1). Ex. using m=3, the bias = 23-1 = 4. Thus the series 0,1,2,3,4,5,6,7 becomes -4,-3,-2,1,0,1,2,3. Therefore, the true exponent -4 is represented by 0 in the bias form and -3 by +1, etc. zero is represented by 0.0 … x 20. Ex. if n = 1010.1111, we normalize it to 0.10101111 x 24. The true exponent is +4. Using the 32-bit standard and a bias of 2m-1-1 = 28-1-1 = 127, the true exponent (+4) is stored as a biased exponent of 4+127 = 131, or 10000011 in binary. Thus we have 0|10000011|01011110000000000000000 Notice that the first 1 in the normalized fraction is omitted. The biased exponent representation is also called excess n, where n is 2m-1-1 (or 2m-1).